0

My math is rusty, but I'll do my best to make this legible:

Let there be numbers $x, y$ where $x>y$.

Let there be a finite multiset $A$ containing at least 1 element and $\exists z \in A$ where $z \geq x$.

We'll also define multiset $B$ as $A \cup \{x\}$ and multiset $C$ as $A \cup \{y\}$ .

We need as simple an example as possible of a function $f$ for which $f(A)>f(B)>f(C)$.

At the same time, for some $w$ where $\not\exists w < q, q\in A$ and multiset $D$ defined as $A \cup \{w\}$, $f(A)<f(D)$.

If certain constraints are necessary for, or would significantly simplify, the answer, feel free to present such a version too.

Thanos Maravel
  • 101
  • 1
    This is not clear at all. Are all these sets (or multisets) of numbers? And what does, say, $f(A)>f(B)$ mean? I'd have guessed that it meant that every number in $f(A)$ was greater than every number in $f(B)$ but $B$ and $C$ intersect so this isn't possible. – lulu Aug 19 '20 at 00:01
  • 1
    A, B and C are multisets, x, y and z are numbers. Applying functions to multisets is where my notation knowledge really gave up on me - they are supposed to be functions applied to the multiset resulting in a single number - for example, multiplying all of them to find a single product (though that will obviously not work for the problem) – Thanos Maravel Aug 19 '20 at 00:05
  • 1
    If it's not what @lulu says but it is literally a function of a set the the inverse size of the set works. – cgss Aug 19 '20 at 00:07
  • 1
    While defining outcomes doesn't give me a function, an inverse size of the set does work for the problem as worded... but not for the original problem, since I neglected a constraint: a new element that is sufficiently bigger than the set's maximum should have the opposite effect on the function. Sorry! And I realize I have poorly worded that extra constraint, too... – Thanos Maravel Aug 19 '20 at 00:14
  • 1
    I don't understand your 6th (out of 7) line. What is big $F$ to small $f$, is it an antiderivative? What do you mean by "for some $w$ where $\not\exists w\dots$? I don't understand that sentence. – Vepir Aug 23 '20 at 20:39
  • 1
    The big F was a typo cause I'm a dummy. – Thanos Maravel Aug 25 '20 at 00:08
  • The goal is to have a function of a set (unspecified set) that, when an element is added to the set that is not (significantly) greater than the greater element of the set, results in a lower value. However, a (significantly) greater value added to the set results in a higher value of the function. Does this make sense? – Thanos Maravel Aug 25 '20 at 00:15
  • Are the elements real numbers? What does "significantly greater" mean? Do you mean something like if $x-y\ge \alpha$ then $x$ is "significantly greater than $y$" but else (in every other case, i.e. $x-y\lt \alpha$) we have "x is not significantly greater than $y$"? For example, say $\alpha=2$, then for example, can you tell me which is greater, $f({1,4})$ or $f({2,3})$ ? – Vepir Aug 25 '20 at 13:57
  • The elements are real numbers. Your definition with $a$ certainly works, but it just means to say that there is leeway on the 'big new number' clause. I don't know which function of the two would result in the greater number, but $f({3,4,6})>f({3,4})>f({2,3,4})$ (for an $a = 2$). – Thanos Maravel Aug 26 '20 at 17:31

0 Answers0